We can be tricky to find the circumference of an ellipse. That is why elliptic integrals were developed.
Anyway, here is a way. But it ain't pretty. Why couldn't you have wanted area, or anything but circumference.
If we solve the ellipse equation for y, we get \(\displaystyle y=\frac{b}{a}\sqrt{a^{2}-x^{2}}\)
Because the ellipse is symmetric wrt the x and y axes, we know the circumference is 4 times the arc length of
\(\displaystyle y=\frac{b}{a}\sqrt{a^{2}-x^{2}}\) in the first quadrant.
The function y in differentiable for all x in the interval [0,a] except at x=a.
Therefore, the circumference is given by an improper integral. Let's call it C:
\(\displaystyle C=4\int_{0}^{a}\sqrt{1+(y')^{2}}dx=4\int_{0}^{a}\sqrt{1+\frac{b^{2}x^{2}}{a^{2}(a^{2}-x^{2})}}dx\)
Use the trig sub \(\displaystyle x=a\cdot sin{\theta}, \;\ dx=a\cdot cos{\theta}d{\theta}\)
\(\displaystyle C=4\int_{0}^{\frac{\pi}{2}}\sqrt{1+\frac{b^{2}sin^{2}{\theta}}{a^{2}cos^{2}{\theta}}}(a\cdot cos{\theta})d{\theta}\)
\(\displaystyle 4\int_{0}^{\frac{\pi}{2}}\sqrt{a^{2}cos^{2}{\theta}+b^{2}sin^{2}{\theta}}d{\theta}\)
\(\displaystyle =4\int_{0}^{\frac{\pi}{2}}\sqrt{a^{2}(1-sin^{2}{\theta})+b^{2}sin^{2}{\theta}}d{\theta}\)
\(\displaystyle =4\int_{0}^{\frac{\pi}{2}}\sqrt{a^{2}-(a^{2}-b^{2})sin^{2}{\theta}}d{\theta}\)
Because
*\(\displaystyle e^{2}=\frac{c^{2}}{a^{2}}=\frac{(a^{2}-b^{2})}{a^{2}}\), we can write as:
\(\displaystyle C=4a\int_{0}^{\frac{\pi}{2}}\sqrt{1-e^{2}sin^{2}{\theta}}d{\theta}\)
*That is not the logarithmic e, but eccentricity of the ellipse 'e'.
Now, we have a elliptic integral of the second kind. That is another matter.
There are entire texts dealing with these. They are beyond the scope of elementary calculus such as parts, substitution, trig, etc.
Given actual values for a and b, one generally looks them up in a table of elliptic integrals \(\displaystyle E(k,{\phi})\)
But, we could expand a power series in \(\displaystyle e^{2}sin^{2}({\phi})\) for small e and integrate term by term.
Given the general nature of your ellipse, I would stop here.
